\documentclass{article}
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\usepackage[mathcal]{euscript}

\usepackage[hidelinks, colorlinks=true, linkcolor=blue]{hyperref} %这个负责交叉引用, [取消红框]
\title{Thought Questions: \\The Geometric/Abstract Definition of Tangent Space}
\author{Jiaqi Tang}
\date{\today}
\linespread{1.25}
\newcounter{problemname}
\newenvironment{problem}{\stepcounter{problemname}\par\noindent\textbf{Problem \arabic{problemname}. }}{\par}
\newenvironment{solution}{\par\noindent\textbf{Solution. }}{\par}
\newenvironment{note}{\par\noindent\textbf{The Note of Problem\arabic{problemname}. }}{\par}

\begin{document}
\maketitle
\begin{abstract}
	These are important exercises for understanding differential and tangent spaces. 
\end{abstract}
Given an open set $\Omega\subseteq \mathbb R^n$ and $p\in \Omega$, let $\mathcal C(p)$ denote the set of all $C^1$ curves through $p$. By definition:
	\begin{align*}
		\mathcal C(p):=\Big\lbrace \gamma\colon I\to \Omega\Big|\text{$I\subseteq \mathbb R$ is an open interval containing $0$, $\gamma(0)=p$, $\gamma\in C^1$}\Big\rbrace ,
	\end{align*}
	we define an equivalence relation on $\mathcal{C}(p)$ (for $\gamma_1, \gamma_2 \in \mathcal{C}(p)$):
\begin{align*}
	\gamma_1 \sim \gamma_2 \iff \gamma_1^\prime(0) = \gamma_2^\prime(0).
\end{align*}
Consider the quotient set $\mathcal{C}(p)/\sim$ under this relation:
\begin{align*}
	\mathcal{C}(p)/\sim = \left\lbrace [\gamma] \mid \gamma \in \mathcal{C}(p) \right\rbrace ,
\end{align*}
where $[\gamma] = [\gamma']$ if $\gamma \sim \gamma'$. 
Geometrically, curves with identical tangent vectors at $p$ are identified, justifying the name ``tangent space'' for this set.

\begin{problem}%1
	Prove the map
	\begin{align*}
		\iota_p\colon \mathcal C(p)/\sim\to \mathbb R^n,\quad [\gamma]\mapsto \gamma^\prime(0)
	\end{align*}
	is well-defined and a bijection. 
\end{problem}

\begin{solution}
	If $[\gamma_1]=[\gamma_2]$, then $\gamma_1\sim\gamma_2$, so $\gamma_1^\prime(0)=\gamma_2^\prime(0)$. Thus, $\iota_p([\gamma_1])=\gamma_1^\prime(0)=\gamma_2^\prime(0)=\iota_p([\gamma_2])$, so $\iota_p$ is well-defined. 
	
	We prove it is injective and surjective:
	\begin{itemize}
		\item \textbf{Injective. }If \(\iota_p([\gamma_1]) = \iota_p([\gamma_2])\), then \(\gamma_1^\prime(0) = \gamma_2^\prime(0)\), so \(\gamma_1 \sim \gamma_2\), hence \([\gamma_1] = [\gamma_2]\).
		\item \textbf{Surjective. }For any $x\in\mathbb R^n$, we can easily get $\iota_p^{-1}(x)=[\gamma(t)=p+xt+o(t)]$
	\end{itemize}
\end{solution}

\begin{problem}%2
	Prove $\mathcal C(p)/\sim$ can be given a $\mathbb R$-linear structure making $\iota_p$ a linear isomorphism. Denote this linear space (originally a set of curve equivalence classes) by $\operatorname T_p\Omega$
\end{problem}

\begin{solution}
	First, we define addition: For $[\gamma_1], [\gamma_2]\in\mathcal C(p)/\sim$, let $[\gamma_1]+[\gamma_2]:=[\gamma_1+\gamma_2]$.
	
	Then we define scalar multiplication: For $c\in\mathbb R, [\gamma]\in\mathcal C(p)/\sim$, let $c\cdot [\gamma]:=[c\cdot \gamma]$.
	
	Now $\operatorname T_p\Omega=\mathcal C(p)/\sim$ is an $\mathbb R$-linear space. We have already know that the map $\iota_p$ is a bijection, hence we just need to show that $\iota_p$ is also a linear map:
	\begin{itemize}
		\item $\iota_p([\gamma_1]+[\gamma_2])=\iota_p([\gamma_1+\gamma_2])=(\gamma_1+\gamma_2)^\prime(0)=(\gamma_1)^\prime(0)+(\gamma_2)^\prime(0)=\iota_p([\gamma_1])+\iota_p([\gamma_2])$.
		\item $\iota_p(c\cdot [\gamma])=\iota_p([c\cdot \gamma])=(c\cdot\gamma)^\prime(0)=c\cdot\gamma^\prime(0)=c\cdot\iota_p([\gamma])$.
	\end{itemize} 
	Therefore, $\iota_p$ is a linear isomorphism. 
\end{solution}

\begin{problem}%3
	Let $\Omega^\prime=\mathbb R^m$ and $f\colon \Omega \to \Omega^\prime$ be $C^1$ (i.e., its differential is continuous), with $p^\prime=f(p)$.
	Prove the map
	\begin{align*}
		f_{\sharp,p}\colon\mathcal C(p)\to \mathcal C(p^\prime),\quad \gamma\mapsto f\circ \gamma
	\end{align*}
	is well-defined. Geometrically, composing with $f$ pushes forward curves through $p$ in $\Omega$ to curves through $p^\prime$ in $\Omega^\prime$.
\end{problem}

\begin{solution}
	$(f\circ \gamma)(0)=f(p)=p^\prime$, $f\in C^1$, $\gamma\in C^1$, hence $f\circ \gamma\in\mathcal C(p^\prime)$.
	
	For any $\gamma_1=\gamma_2\in\mathcal C(p)$, we have $f_{\sharp,p}(\gamma_1)= f\circ \gamma_1=f\circ \gamma_2=f_{\sharp,p}(\gamma_2)$, so $f_{\sharp,p}$ is well-defined. 
\end{solution}

\begin{problem}%4
	Prove the map
	\begin{align*}
		f_{*p}\colon \operatorname T_p\Omega\to \operatorname T_{p^\prime}\Omega^\prime,\quad [\gamma]\mapsto [f_{\sharp,p}(\gamma)]
	\end{align*}
	is a well-defined linear map (called the \textbf{tangent map}). 
\end{problem}

\begin{solution}
	For any $[\gamma_1]=[\gamma_2]$, then $\gamma_1\sim\gamma_2$, so $\gamma_1^\prime(0)=\gamma_2^\prime(0)$. Thus, 
	\begin{align*}
		f_{*p}([\gamma_1])=[f\circ\gamma_1]=[f\circ \gamma_2]=f_{*p}([\gamma_2])
	\end{align*}
	The second equality is holds because we have $(f\circ\gamma_1)^\prime(0)=\operatorname d f(p)(\gamma_1^\prime(0))=\operatorname d f(p)(\gamma_2^\prime(0))=(f\circ\gamma_2)^\prime(0)$ for any $C^1$ map $f\colon \Omega\to \Omega^\prime$. So $f_{*p}$ is well-defined. 
	
	Now we prove its linearity: 
	\begin{itemize}
		\item $f_{*p}([\gamma_1]+[\gamma_2])=[f\circ(\gamma_1+\gamma_2)]=[f\circ \gamma_1+f\circ\gamma_2]=[f\circ \gamma_1]+[f\circ\gamma_2]=f_{*p}([\gamma_1])+f_{*p}([\gamma_2])$.
		\item $f_{*p}(c\cdot[\gamma])=f_{*p}([c\cdot\gamma])=[f\circ(c\cdot\gamma)]=[c\cdot (f\circ \gamma)]=c\cdot[f\circ\gamma]=c\cdot f_{*p}([\gamma])$.
	\end{itemize}
	So $f_{*p}$ is a well-defined linear map. 
\end{solution}

\begin{problem}%5
	Consider the curve on $\Omega$:
\begin{align*}
	l_k\colon (-1, 1) \to \Omega,\quad t \mapsto p + \underbrace{(0, 0, \cdots, 0, t, 0, \cdots, 0)}_{\text{only the $k$-th component is non-zero}}.
\end{align*}
Prove $\iota_p([l_k]) = \dfrac{\partial}{\partial x_k}$, where $\lbrace x_1, \dots, x_n\rbrace $ is the coordinate system on $\mathbb{R}^n$.
\end{problem}

\begin{solution}
	Notice that $\iota_p([l_k])=l_k^\prime(0)=l_k^\prime(t)|_{t=0}=\left.\dfrac{\partial}{\partial x_k}\right|_{t=0}=\dfrac{\partial}{\partial x_k}.$
\end{solution}

\begin{problem}%6
	Consider the curve on $\Omega^\prime$:
\begin{align*}
	l_{k^\prime}^\prime\colon (-1, 1) \to \Omega^\prime,\quad t \mapsto p + \underbrace{(0, 0, \cdots, 0, t, 0, \cdots, 0)}_{\text{only the $k$-th component is non-zero}}.
\end{align*}
Express $f_{*p}([l_k])$ using the partial derivatives of $f$ and $[l_{k^\prime}^\prime]$.
\end{problem}

\begin{solution}
	$f_{*p}([l_k])=[f\circ l_k]=\sum\limits_{k^\prime=1}^m\Big (\nabla_{\frac{\partial}{\partial x_k}}f_{k^\prime}\Big)(p)[l^\prime_{k^\prime}]$
	
	The second equality is holds because
	\begin{align*}
		(f\circ l_k)^\prime(0)=\operatorname df(p)\left(\dfrac{\partial}{\partial x_k}\right)=\Big(\nabla_{\frac{\partial}{\partial x_k}}f\Big)(p)=\sum\limits_{k^\prime=1}^m\Big(\nabla_{\frac{\partial}{\partial x_k}}f_{k^\prime}\Big)(p)\left(\dfrac{\partial}{\partial x_{k^\prime}}\right).
	\end{align*}
\end{solution}

\begin{problem}%7
	Prove the commutative diagram: 
	\begin{center}
		\begin{tikzcd}
			\operatorname T_p\Omega \arrow[r,"\iota_p","\cong"']\arrow[d,"f_{*p}"]&\mathbb R^n\arrow[d,"\operatorname df(p)"]\\
			\operatorname T_{p^\prime}\Omega^\prime\arrow[r,"\iota_{p^\prime}","\cong"'] & \mathbb R^m
		\end{tikzcd}
	\end{center}
For any $[\gamma]\in \operatorname T_p\Omega$, we have 
 \begin{align*}
 	f_{*p}([\gamma])=\iota_{p^\prime}^{-1}\Big(\operatorname df(p)\big(\iota_p([\gamma])\big)\Big).
 \end{align*}
In this sense, $f_{*p}$ coincides with $\operatorname df(p)$.
\end{problem}

\begin{solution}
	For any $[\gamma]\in\operatorname T_p\Omega$, we have 
	\begin{align*}
		\iota_{p^\prime}(f_{*p}([\gamma]))=\iota_{p^\prime}([f\circ \gamma])=(f\circ \gamma)^\prime(0)=\operatorname df(p)(\gamma^\prime(0))=\operatorname df(p)(\iota_p([\gamma])),
	\end{align*}
	which shows that $\iota_{p^\prime}\circ f_{*p}=\operatorname df(p)\circ \iota_p$. 
\end{solution}

\begin{problem}%8
	For non-zero $[\gamma_1], [\gamma_2] \in\operatorname T_p\Omega$, define their angle by:
\begin{align*}
	\cos\big(\angle([\gamma_1], [\gamma_2])\big) = \frac{\gamma_1^\prime(0) \cdot \gamma_2^\prime(0)}{\|\gamma_1^\prime(0)\| \, \|\gamma_2^\prime(0)\|},
\end{align*}
with the angle in $[0, \pi)$. Consider the inversion map:
\begin{align*}
	f\colon\mathbb{R}^n \setminus \lbrace 0\rbrace  \to \mathbb{R}^n \setminus \lbrace 0\rbrace ,\quad x \mapsto \dfrac{x}{\|x\|^2}.
\end{align*}
Prove $f$ is $C^1$ and for all $p \in \mathbb{R}^n \setminus \lbrace 0\rbrace $, $f_{*p}\colon\operatorname T_p(\mathbb{R}^n \setminus \lbrace 0\rbrace ) \to\operatorname T_{f(p)}(\mathbb{R}^n \setminus \lbrace 0\rbrace )$ preserves angles.
\end{problem}

\begin{solution}
	$f=(f_1,f_2,\cdots,f_n)$. For all $1\leqslant i,k\leqslant n$, we have 
	\begin{align*}
		\nabla_{\frac{\partial}{\partial x_k}} f_i=\big(\delta_i^k\|x\|^2-2x_ix_k\big)\big/{\|x\|^4}\in C(\mathbb R^n\setminus\lbrace 0\rbrace ). 
	\end{align*}
	So $f\in C^1$. For non-zero $[\gamma_1],[\gamma_2]\in\operatorname T_p(\mathbb R^n\setminus\lbrace 0\rbrace )$, we denote $\gamma_1^\prime(0), \gamma_2^\prime(0)$ by $v_1,v_2$ respectively. 
	\begin{align}\label{1}
		\cos\big(\angle(f_{*p}([\gamma_1]),f_{*p}([\gamma_2])\big)=\dfrac{(f\circ\gamma_1)^\prime(0)\cdot(f\circ\gamma_2)^\prime(0)}{\|(f\circ\gamma_1)^\prime(0)\|\|(f\circ\gamma_2)^\prime(0)\|}=\dfrac{\operatorname df(p)(v_1)\cdot\operatorname df(p)(v_2)}{\|\operatorname df(p)(v_1)\|\|\operatorname df(p)(v_2)\|}.
	\end{align}
	Denote $\operatorname df(p)(v_1),\operatorname df(p)(v_2)$ by $w_1,w_2$ respectively. 
%	\begin{align*}
%	\operatorname{Jac}(f)|_{p}=\left(\dfrac{\delta_i^j\|p\|^2-2p_ip_j}{\|p\|^4}\right)_{\substack{1\leqslant i\leqslant n\\1\leqslant j\leqslant n}}.
%	\end{align*}

	Notice that $w_1\cdot w_2=\operatorname{det}(\mathbf J^\top \mathbf J)v_1\cdot v_2$, $\|w_i\|=\sqrt{w_i\cdot w_i}=\sqrt{\operatorname{det}(\mathbf J^\top \mathbf J)}\|v_i\|,i=1,2$. Hence,
	\begin{align*}
		\eqref{1}=\dfrac{w_1\cdot w_2}{\|w_1\|\|w_2\|}=\dfrac{\operatorname{det}(\mathbf J^\top \mathbf J)v_1\cdot v_2}{\sqrt{\operatorname{det}(\mathbf J^\top \mathbf J)}\|v_1\|\sqrt{\operatorname{det}(\mathbf J^\top \mathbf J)}\|v_2\|}=\dfrac{v_1\cdot v_2}{\|v_1\|\|v_2\|}=\cos\big(\angle([\gamma_1], [\gamma_2])\big).
	\end{align*}
\end{solution}

\begin{flushbottom} 
\noindent ------------------------

Pure mathematics is, in its way, the poetry of logical ideas.
\flushright{ -- Albert Einstein}

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\end{flushbottom}







\end{document}